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Exponents and Logarithms
In math, every operation has another operation that allows you to work backwards—addition has subtraction, multiplication has division and exponents have logarithms. Logarithms are used to determine what exponent a given base would require in order to produce a certain number. For further clarification assume the following generic expression for exponents:
bx y Equation 1 where, b is called the “base”, x is the exponent and y is that certain number. With that in mind, we have the following as a generic expression for logarithm: log by x Equation 2
where, all the variables are given the same label as above (b is base, x is exponent and y is that certain number).
Using those first two equations (Equation 1 and 2) you can substitute x in Equation 1 with the statement in Equation 2 to find this rule:
b log by y Equation 3 Notice how the subscript of “log” is the same as the base? If both of those are equal, then that expression will always hold true. This becomes important later on.
Examples usingb log by y
- 4 log^41717 2.^10 log^2121
- e ln^100100 No calculators were needed to figure those out. No real thought. All you do is realize you are in the form of (^) b log by y and then use that to your advantage. You may notice
that Examples 2 and 3 may look weird. “What is ‘ln’?” and “Why isn’t there a b by the log?” Well, let me explain.
There are two special logarithms that are used so frequently they’ve been given special notation. When you simply see “log” it is assumed to be of base 10 (our favorite of all the bases). When you see “ln” that is simply a “log” with a base of e (loge) which is
bx y log by x b base base x exponent exponent y that “certain number”
that “certain number”
also known as “natural log”. Don’t know what “e” is? Check out the “What is e?” note below.
When dealing with logarithms there are eight (8) special rules (or properties) that will make solving logarithmic problems easier. The rules are assuming M, N and b are all positive, non-zero numbers—there is no way to get the log of a negative number or zero. For example, log 2 4 is not possible. Feel free to grab your calculator if you don’t believe me.
An important thing to remember is this works both left to right and right to left. In other words, log b M log bN log bMN is just as legitimate as log b MN log bM log bN ).
Examples using the aforementioned rules:
- log 233 log 211 log 23
- log 5 log 10 log 2 (because 10 divided by 2 is 5—I could have just as easily did log 20 log 4 or any other combination of numbers that, when divided, gives me a quotient of 5)
- log 4 log 7 7
log 4 3 3 3
- log 6 x^22 log x
- log 216 4 log 22 4 ( 1 ) 4 ( remember log bbalways equals 1 )
Some of these you can cleverly manipulate to use different rules. Who’s to say you need to use 11 and 3 for Example 1? You could have done log 233 log 299 log 23 where 99 3 33 just as easily.
One last concept to have down and we should be good to go! What if you have something like log 27? Remember, that’s asking “Two (2) to what power is 7?” Well, there’s something you can do to make it easier: Change of Base. To do change of base, all you do is follow this formula:
Rules
- (^) log (^) b 1 0 2. log (^) bb 1
- (^) log bbx x 4. b log bx x
- (^) log b MN log bM log bN 6. M N N
M
log b log b log b
- (^) log b M p p log bM 8. (^) If log b x log by , then x y
What is e?
It is a special number, like π (3.1415…), called a transcendental number. A transcendental number is an irrational number (a number that can’t be written into a fraction) and is not a solution of a non-zero polynomial. Do you need to know this to work these problems? No. All you need to know is that e (which is 2.71828…) represents a number (just like π). ex When x = ex^ = x = 0 e^0 = x = 1 e^1 =2.71828… x = 2 e^2 =7.38906…
describes this is the following (where A is the amount of radioactive material present at time t and A 0 is the amount initially present): 5700 0 2 A A t So, let’s take a look at some examples using what we’ve learned so far. Another fair warning, though: review the following key exponent concepts (it will make things so much easier if you understand these).
So, why don’t we work some examples? Sounds like fun, right?
Example Answer
Solve for x to four decimal places
- 7 x 12
7 x 12 (^) log 7 x log 12 take the log of both sides x log 7 log (^12) using Rule 7 from above
log 7
log 12 x divide both sides by log 7 to get x alone
x 1. 2770 Solve for x to four decimal places
- 5 2 x^^1125 x
5 2 x^^1125 x log 52 x^^1 log 125 x 125 53 rewrite 125 as to have a base of 5 and plug it back into the equation 2 x 1 log 5 x log 53 2 x 1 log 5 3 x log 5 2 x log 5 log 5 3 x log 5 after distributing log into 2x+ 2 x log 5 3 x log 5 log 5 subtract 3xlog 5 from both sides, bringing it over to the left hand side, then subtract log5 from both sides, bringing it over to the right hand side. x log 5 log 5 after subtracting the terms on the left hand side
log 5
x log^5 dividing both sides by – log
x 1 remember, if the numerator is the same as the denominator, they cancel each other out and leave a 1 in their place
Solve for x to four decimal places
- 3 x^3 3 x 84 (Hint: 3 x^3 3 x 33 )
3 x^3 3 x 84 rewrite after applying the hint 3 x 33 3 x (^84) notice how both terms on the left side have a 3x^ in them? Let’s factor that out. 3 x 33 1 84 3 x 27 1 84 3 x 28 84
3 x^84
3 x 3 when the bases are the same on both the left hand side and the right hand side (in this case the bases are both 3) then the exponents have to be equal, therefore: x 1 there really was no need to take the log of both sides (though if you did, you will still get the same answer) Solve 4.log( 3 x 5 ) log( 2 x 6 ) 0
log( 3 x 5 ) log( 2 x 6 ) (^0) add log(2x+6) to both sides log( 3 x 5 ) log( 2 x 6 ) now, put these two equations as exponents to the same base (chose the same base that the log is worked in) 10 log(^3 x^5 ) 10 log(^2 x^6 ) remember 10 is the understood base of a logarithm if there is no subscript attached to the word “log” in the equation 3 x 5 2 x 6 from Rule 4 above 3 x 2 x 6 5 using basic Algebra concepts (subtracting both 2x and 5 from both sides) x 1 not quite done yet—have to plug this back in to the original equation to make sure we don’t break any logarithm rules (namely the rule about taking the log of a negative number) Another way of looking at this… log( 3 x 5 ) log( 2 x 6 ) 0 add log(2x+6) to both sides log( 3 x 5 ) log( 2 x 6 ) if both sides have a log with the same base (remember, this one is base 10 on both sides) we can just set what’s inside the parenthesis to be equal 3 x 5 2 x 6 3 x 2 x 6 5 using basic Algebra concepts (subtracting both 2x and 5 from both sides) x 1 as you can see, we still get the same answer Checking… log( 3 ( 1 ) 5 ) log( 2 ( 1 ) 6 ) 0 log( 8 ) log( 8 ) 0 This satisfies the rules of logarithms and, therefore, x 1 is a valid solution of
